A capacitor C is fully charged with voltage V0. After disconnecting the voltage source, it is connected in parallel with another uncharged capacitor of capacitance ${C \over 2}$. The energy loss in the process after the charge is distributed between the two capacitors is :
Solution
Heat loss = ${1 \over 2}\left( {{{{C_1}{C_2}} \over {{C_1} + {C_2}}}} \right)V_0^2$
<br><br>= $${1 \over 2}\left( {{{C \times {C \over 2}} \over {C + {C \over 2}}}} \right)V_0^2$$
<br><br>= ${1 \over 6}CV_0^2$
About this question
Subject: Physics · Chapter: Electrostatics · Topic: Electric Potential and Capacitance
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